The Signed Domination Number of Cartesian Product of Two Paths

Let G be a finite connected simple graph with vertex set V(G) and edge set E(G). A function ( ) { } : 1,1 f V G → − is a signed dominating function if for every vertex ( ) v V G ∈ , the closed neighborhood of v contains more vertices with function values 1 than with −1. The signed domination number ( ) s G γ of G is the minimum weight of a signed dominating function on G. In this paper, we calculate The signed domination numbers of the Cartesian product of two paths P m and P n for m = 3, 4, 5 and arbitrary n.


Introduction
Let G be a finite simple connected graph with vertex set V(G) and edge set E(G). The . The function f is a signed dominating function if for every vertex v V ∈ , the closed neighborhood of v contains more vertices with function value 1 than with −1.  [4], in [5] Haas and Wexler had found the signed domination number of P 2 × P n and P 2 × C n . In [6] Hosseini gave a lower and upper bound for the signed domination number for any graph.
We consider when we represent the P m × P n graph to find the signed dominating function that the black circles refer to the graph vertices which weight 1, signed dominating function of the P m × P n graph and, ( )

Main Results
In this paper we will show three theorems to find the signed domination number of Cartesian product of P m × P n . Theorem 2.1. Let n be a positive integer: Proof: Case n ≡ 0 (mod 3) Let f be a signed dominating function of (P 3 × P n ), then for any j were 2 ≤ j ≤ n − 1, then We discuss the following cases: Case a. |B j | = 2 ( Figure 1) We notices that the first and last columns can't include more than one vertex of the B set vertices. But in the case 2 ≤ j ≤ n − 1 and |B j | = 2, the vertices (1, j) and (3, j) belong to the B set vertices and all the

3, j B
∈ then all the vertices (2, j − 1), (3, j − 1), (2, j + 1) and (3, j + 1) belong to the A set, and one of the vertices (1, j − 1) or (1, j + 1) at most belong to the B set vertices.  Case c. |B j | = 0 ( Figure 3) When the j th column doesn't include any one of the B set vertices, it is possible that the vertices (1, j + 1) and (3, j + 1) belong to the B set provided that the j + 1 th column isn't the last column then all the j + 2 th column vertices belong to the A set, or the tow vertices (1, j − 1) and (3, j − 1) belong to the B set provided that the j − 1 th column isn't the first one, and all the j − 2 th column vertices belong to the A set.
Whereas the j + 1 th column includes one of the B set vertices, then the j + 2 th column will include one of the B set vertices at most. Also if the j − 1 th column includes one of the B set then the j − 2 will include one of the B set vertices at most.
We conclude from the previous cases that if 2 ≤ j ≤ n -1, then And all three successive columns include two vertices at most weighted with −1, so seven vertices at least is weighted the weight 1, consequently: To find the upper bound of the signed domination number of (P 3 × P n ) graph, Figure 4).
If B is the previously defined set and represents the vertices have the weight −1, then every one of the P 3 × P n graph vertices achieves the signed dominating function, and Case n ≡ 1 (mod 3) ( Figure 5) If we add a column to the previous graph in case  then one vertex at most can have the weight −1, so, when we add that vertex to the B set this makes f a signed dominating function, so: Consequently:    In this case we add to two columns to the graph so, the numeral to the vertices in the B set at any two successive columns is less or equals 2, so the signed domination number will increase of 2 than the signed domination number in case of ( ) 0 mod 3 n ≡ . If we add the vertices (n, 2) and (n − 1, 0) to B set then f remains a signed dominating function of the graph, and Consequently, the domination number will be: Proof: Let f be a signed domination function of the (P 4 × P n ) graph. And let A, B, K j , A j and B j are the previously defined sets, Whatever j is then we notice that B j ≤ 2 so, discuss the following cases: Case a. |B j | = 2. Then: In this case all the j + 1 th column vertices belong to the A set vertices so, the two remained vertices of the j th column (Figure 7): In this case one vertex at most j + 1 th or j − 1 th column vertices can belong to the B set vertices, either Figure 8).
The vertices (1, j), (2, j) can't belong to the B set at the same time, neither the vertices (3, j), (4, j) because both of the vertices (1, j), (4, j) from the third degree and can't connect with any one of the B set vertices.
Case b. |B j | = 1: then we discuss the following cases: ∈ then one of the j + 1 th column vertices at most can be from the B set vertices (Figure 9) ∈ then two of the j + 1 th column vertices at most can be from the B set, and in this case all the j + 2 th column vertices are from the A set vertices, and one vertex of the j − 1 th column vertices at most can belong to the B set vertices, in this case We noticed that if the B set vertices are the vertices which have the weight −1 of the P 4 × P n graph, every one of the graph vertices achieves the signed dominating function so, the signed domination number of the P 4 × P n graph will be:       Proof: Let f be a signed domination function of the P 5 × P n graph. And A, B, K j , A j and B j are the previously defined sets, then whatever 1 ≤ j ≤ n -4, then:

4, j B
∈ then the j + 1 th column doesn't include any one of the B set vertices. The j + 2 th column include three of the B set vertices, the j + 3 th column doesn't include any one of the B set vertices. Then every two successive columns include three vertices of the B set. And every ten successive columns include fifteen vertices of the B set.
b.4. If (2, j) and (3, j) ∈ B then (4, j + 1) or (5, j + 1) ∈ B, and the j + 2 th column include two vertices of the B set vertices at most, And the j + 3 th column include one vertex at most. And the j + 4 th column include only two vertices.
b.5. If (2, j) and (4, j) ∈ B then (2, j + 1), (4, j + 1) ∈ B. And the j + 2 th column doesn't include any one of the B set vertices (1, j + 3), (3,  Case c. |B j | = 1 (Figure 15) c.1. If (1, j) ∈ B or (2, j) ∈ B or (4, j) ∈ B or (5, j) ∈ B then the j + 1 th column include two of the B set vertices and the j + 2 th column include one vertex at most. And the j + 3 th column include two of the B set vertices at most. In this case the j + 4 th column include one of the B set vertices at most. c.2. If (3, j) ∈ B then (1, j+1)∈ B, (3, j + 1) ∈ B, (5, j + 1) ∈ B. And the j + 2 th column doesn't include any one of the B set vertices. And the vertices (2, j + 3), (4, j + 3), (2, j + 4) and (4, j + 4) belong to the B set vertices and the other cases are repeated. Whatever 1 ≤ j ≤ n -4, then s n s n s n P P n P P n P n n n P γ γ γ Let's defined:  Figure 16). Open Journal of Discrete Mathematics  We noticed that if the B set vertices are the vertices which have the weight −1 of the graph P 5 × P n then every one of the graph vertices achieve the signed domination function then the signed domination number of the P 5 × P n graph will be: ( ) But it proves easily, that the first and second columns of P 5 × P n , have at most three vertices of the B set vertices. As well the vertices (1, 1) and (3, 1) they cannot belong to set B in the same time, then if we delete the vertex (3, 1) of B set, then the signed domination number of the P 5 × P n will be:

Conclusion
In this paper, we studied The signed domination numbers of the Cartesian product of two paths P m and P n for m = 3, 4, 5 and arbitrary n. we will work to find the signed domination numbers of the Cartesian product of two paths P m and P n for arbitraries m and n.